A particle starts from rest and moves in a straight line. The velocity of the particle at time t s after the start is v m s⁻¹, where v = −0.01t³ + 0.22t² – 0.4t.
- (i)Find the two positive values of t for which the particle is instantaneously at rest.[2]
- (ii)Find the time at which the acceleration of the particle is greatest.[3]
- (iii)Find the distance travelled by the particle while its velocity is positive.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [M1] -0.01t(t² - 22t + 40) = 0 or -0.01t(t - 20)(t - 2) = 0 Attempting to solve v = 0 for t for a solvable quadratic using factors or quadratic formula and obtaining two non-zero solutions
- [A1] t = 2 or t = 20
- [M1] a = -0.03t² + 0.44t - 0.4 For differentiation
- [M1] a is greatest (maximum) when 0.44 - 0.06t = 0 For differentiation or finding values of t = t₁ and t = t₂ where a = 0 and using t = ½(t₁ + t₂) or completing the square or other method to find maximum value
- [A1] Max acceleration when t = 7.33 Allow t = 22/3
- [*M1] ∫(-0.01t³ + 0.22t² – 0.4t) dt For using integration.
- [A1] s(t) = -0.01/4 t⁴ + 0.22/3 t³ - 0.2t² Correct Integration Allow + C included
- [DM1] s(20) - s(2) Limits 2 and 20 used correctly Dependent on previous M1 having been scored
- [A1] Distance = 107 m Distance = 2673/25 = 106.92
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